Ohm's law, series and parallel circuits

Physics · Senior school physics and first-year engineering · about 10 min

Every electrical circuit, from a torch to a phone charger, follows Ohm's law. Combine it with two rules for resistors in series and in parallel and you can work out the current and voltage anywhere in a simple circuit.

The key idea

V = I R

Resistors in series add: $R = R_1 + R_2 + \dots$

Resistors in parallel combine as $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$ (for two, $R = \dfrac{R_1 R_2}{R_1 + R_2}$ ).

Series versus parallel

Series: one path, so the same current flows through every resistor, and the supply voltage is shared between them.
Parallel: several paths, so each branch has the same voltage, and the current splits, more of it through the smaller resistance.

A parallel combination is always smaller than its smallest resistor: adding a path makes it easier for current to flow.

Two circuit diagrams side by side, in thin black lines with standard zigzag resistor symbols and a standard battery symbol. Left: a single rectangular loop with a battery on the left side and two resistors one after the other along the top. Right: a battery on the left side, and two resistors in separate branches, one above the other, across the same two points. — A series circuit has one path, and a parallel circuit has several.

How to solve a mixed circuit

Collapse parallel groups into single equivalent resistors.
Add the series parts to get the total resistance.
Use $I = V/R$ for the total current from the supply.
Work back out: find the voltage across each part, then each branch current.
Check: currents into a junction equal currents out, and voltages around a loop add up to the supply.

Worked example: a series-parallel circuit

A $12$ V battery drives a $4\,\Omega$ resistor in series with a parallel pair of $6\,\Omega$ and $3\,\Omega$ . Find the total current, and the current through each of the parallel resistors.

A circuit diagram in thin black lines with standard zigzag resistor symbols and a standard battery symbol. A battery on the left side of a rectangular loop. One resistor along the top. After it, on the right side of the loop, two resistors in parallel branches, side by side. A small dark red arrow on the top wire shows the current direction. — The circuit in this example.

Show the solution

Parallel pair:

R_p = \frac{6 \times 3}{6 + 3} = 2\,\Omega

Total resistance: $R = 4 + 2 = 6\,\Omega$

Total current:

I = \frac{12}{6} = 2 \text{ A}

Voltages: across the $4\,\Omega$ resistor, $V = 2 \times 4 = 8$ V; across the parallel pair, $V = 2 \times 2 = 4$ V (and $8 + 4 = 12$ V, which checks).

Branch currents: $I_6 = \frac{4}{6} = 0.67$ A and $I_3 = \frac{4}{3} = 1.33$ A, which add back to $2$ A.

Common mistakes

Adding parallel resistors like series ones

Parallel resistances combine through their reciprocals, and the result is smaller than either.

Using the supply voltage across one resistor in a series chain

In series the supply voltage is shared. Use the current and that resistor: $V = IR$ .

Thinking current gets used up along a wire

Current is the same all the way round a series loop. It's energy (voltage) that gets used up.

Check your understanding

1. What is the combined resistance of $10\,\Omega$ and $10\,\Omega$ in parallel?

2. A $9$ V battery drives a $3\,\Omega$ resistor. What current flows?

3. In a parallel circuit, which quantity is the same across every branch?

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